Compact Hausdorff under both $\mathcal{T}$ and $\mathcal{T'}$

Show that if $X$ is compact Hausdorff under both $\mathcal{T}$ and $\mathcal{T'}$, then either $\mathcal{T}$ and $\mathcal{T'}$ are equal or not comparable.

My solution:

So, we have to show $\mathcal{T}\subset \mathcal{T'}$ and $\mathcal{T}\supset\mathcal{T'}$.

First of all, we know that every subspace of compact Hausdorff is closed. Suppose $\mathcal{T}\subset\mathcal{T'}$. Then, compactness under the topology $\mathcal{T'}$ (trivially) implies compactness under the topology $\mathcal{T}$.

Again, every closed set under the topology $\mathcal{T}$ can be written as finite union of closed sets under the topology $\mathcal{T'}$. Hence $\mathcal{T}\supset\mathcal{T'}$.

Is my argument ok? Thanks.

Edit:

1. Sorry for your inconvenience for the statement "...compact Hausdorff is closed." I meant:- "Every compact subspace of a Hausdorff space is closed"(Munkres Topology Thrm.: 26.3)

2. Sorry for the second part too (The statement: "every closed set under the topology $\mathcal{T}$ can be written as finite union of closed sets under the topology $\mathcal{T'}$"). I wanted to say "Since $\mathcal{T}\subset\mathcal{T'}$, every closed set under the topology $\mathcal{T'}$ can be written as finite union of closed sets under the topology $\mathcal{T}$".

Then let $X-A_1,X-A_2,\dots, X-A_k\in \mathcal{T}$, and $\mathcal{A}_1,\mathcal{A}_2,\dots, \mathcal{A}_k$ are the open covering of $A_1,A_2,\dots,A_k$ respectively, so that $\bigcup_{i=1}^{k}A_i=B$, where $X-B\in \mathcal{T'}$. Then the union of finite subcollections that covers $A_1,A_2,\dots, A_k$ will cover $B$. Then compactness under the topology $\mathcal{T}$ implies compactness under the topology $\mathcal{T'}$. Then $\mathcal{T}\supset\mathcal{T'}$.

• Can you explain how you get the corrected version of the second statement? Also, when you say "$\mathcal{A}_1,\mathcal{A}_2,\ldots,\mathcal{A}_3$ are the covers respectively", what do you mean? Also, what is $B$? – John Griffin Sep 24 '17 at 3:35
• ok, wait, I wrote wrong again! $A_1,A_2,\dots, A_k$ are closed sets and $B$ is also closed set too. $\mathcal{A}_i$ is an open covering of $A_i$. @JohnGriffin can you check now. – topology_001 Sep 24 '17 at 3:52
• @JohnGriffin I edited a little, thanks a lot. – topology_001 Sep 24 '17 at 4:02
• I'm still struggling to follow your new argument. So is $B$ an arbitrary $\mathcal{T}'$-closed set? How do you know you can write such a thing as a finite union of $\mathcal{T}$-closed sets? How does the fact that compactness in $\mathcal{T}$ implies compactness in $\mathcal{T}'$ help to conclude $\mathcal{T}\supset\mathcal{T}'$? – John Griffin Sep 24 '17 at 4:17
• @JohnGriffin thanks, I understood the problem in my argument. – topology_001 Sep 24 '17 at 4:25

"...every subspace of compact Hausdorff is closed." I'm not sure what you mean by this. The space $[0,1]$ is compact and Hausdorff, but its subspace $[0,1/2)$ is not closed.

"...every closed set under the topology $\mathcal{T}$ can be written as finite union of closed sets under the topology $\mathcal{T}'$." Well if $E$ is $\mathcal{T}$-closed, then $E^c$ is $\mathcal{T}$-open, and thus by $\mathcal{T}\subseteq\mathcal{T}'$, we see that $E^c$ is $\mathcal{T}'$-open and consequently $E$ is $\mathcal{T}'$-closed. Thus every $\mathcal{T}$-closed set is $\mathcal{T}'$-closed. However how does this show that $\mathcal{T}\supseteq\mathcal{T}'$?

The easiest way to do this is to assume $\mathcal{T}\subseteq\mathcal{T}'$, and consider the identity map $f:(X,\mathcal{T}')\to(X,\mathcal{T})$ defined by $f(x)=x$. This is continuous by our assumption, and it's a bijection. Since it maps from a compact space into a Hausdorff space, we can conclude that it is a homeomorphism. Therefore $f$ maps open sets to open sets, meaning that if $U$ is $\mathcal{T}'$-open, then $U=f(U)$ is $\mathcal{T}$-open. Therefore $\mathcal{T}=\mathcal{T}'$.

If you aren't familiar with the theorem I used above, you should try proving it independently. It is tremendously useful and its proof relies on the same techniques one would employ to prove your problem without it. Here it is:

If $f:X\to Y$ is a continuous function from a compact space to a Hausdorff space, then $f$ is a closed map (i.e., it maps closed subsets of $X$ to closed subsets of $Y$). Moreover, if $f$ is a bijection, then $f$ is a homeomorphism.

I alluded that you can prove that $\mathcal{T}\subseteq\mathcal{T}'$ implies $\mathcal{T}=\mathcal{T}'$ without using this theorem, and as I suspect this is the route you intended, here's how one would do so. Suppose $U$ is $\mathcal{T}'$-open. Then $U^c$ is $\mathcal{T}'$-closed. Since closed subsets of compact sets are compact, we know that $U^c$ is $\mathcal{T}'$-compact. As you pointed out, $\mathcal{T}'$-compactness implies $\mathcal{T}$-compactness, so $U^c$ is $\mathcal{T}$-compact. Since $\mathcal{T}$ is Hausdorff and compact subsets of Hausdorff spaces are closed, we deduce $U^c$ is $\mathcal{T}$-closed. Therefore $U$ is $\mathcal{T}$-open, which completes the proof.

I find that the following reformulation of this problem is more enlightening.

Let $\tau_C$, $\tau_H$, and $\tau$ be topologies on a set $X$ such that $(X,\tau_C)$ is compact and $(X,\tau_H)$ is Hausdorff.

(i) If $\tau\subseteq\tau_C$, then $(X,\tau)$ is compact.

(ii) If $\tau\supseteq\tau_H$, then $(X,\tau)$ is Hausdorff.

(iii) If $\tau_H \subseteq \tau_C$, then $\tau_C=\tau_H$.

• Very nice! Formulated your way, the relevant properties are intuitively obvious: discarding open sets can't invalidate compactness, while adding open sets can't invalidate the Hausdorff property. – Rob Arthan Sep 23 '17 at 20:13
• Sorry for your inconvenience for the statement "...compact Hausdorff is closed."(I almost fell asleep when I was posting this[ha ha]) I meant:- "Every compact subspace of a Hausdorff space is closed"(Munkres Topology Thrm.: 26.3) – topology_001 Sep 24 '17 at 2:50
• Sorry for the second part too (The statement: "every closed set under the topology $\mathcal{T}$ can be written as finite union of closed sets under the topology $\mathcal{T'}$") I wanted to say "every closed set under the topology $\mathcal{T'}$ can be written as finite union of closed sets under the topology $\mathcal{T}$". – topology_001 Sep 24 '17 at 2:54
• ....Also, I added some explanation. Can you please check it, please! – topology_001 Sep 24 '17 at 3:10
• @topology_001 Haha yeah....it happens. I'm happy to help! – John Griffin Sep 24 '17 at 3:35